A student rides on a bicycle at 5 km/hr and reaches his school 3 minute late. The next day he increased his speed to 7 km/hr and reached school 3 min early. Find the distance between his house and the school.
A student rides on a bicycle at 5 km/hr and reaches his school 3 minute late. The next day he increased his speed to 7 km/hr and reached school 3 min early. Find the distance between his house and the school.
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Let’s denote the distance between the student’s house and the school as \(d\) km.
When the student rides at 5 km/hr, he is 3 minutes late. When he rides at 7 km/hr, he is 3 minutes early. Let’s denote the time it takes to reach the school on time as \(t\) hours.
We can set up two equations based on the information given:
1. When riding at 5 km/hr and being 3 minutes late:
\[ \frac{d}{5} = t + \frac{3}{60} \]
2. When riding at 7 km/hr and being 3 minutes early:
\[ \frac{d}{7} = t – \frac{3}{60} \]
We can solve these two equations simultaneously to find \(d\) and \(t\).
From equation 1:
\[ 60d = 300t + 15 \] (1)
From equation 2:
\[ 60d = 420t – 21 \] (2)
Subtracting equation (2) from equation (1):
\[ 0 = -120t + 36 \]
\[ 120t = 36 \]
\[ t = \frac{36}{120} = \frac{3}{10} \text{ hours} \]
Substituting \(t\) back into either equation (1) or (2) to find \(d\):
\[ 60d = 300 \times \frac{3}{10} + 15 \]
\[ 60d = 90 + 15 \]
\[ 60d = 105 \]
\[ d = \frac{105}{60} = 1.75 \text{ km} \]
Therefore, the distance between the student’s house and the school is 1.75 km.